JEE PYQ: Motion in a Straight Line - Question ID 0e6377476781 (JEE Main 2024)

Core Formula & Concept:

In problems involving motion in a straight line with varying acceleration, the average speed is defined as the total distance traveled divided by the total time taken. Mathematically, $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}.$

When a body accelerates uniformly from rest, its velocity increases linearly with time. The distance covered during this phase can be calculated using the equation of motion: $s = ut + \frac{1}{2} a t^2,$ where $u$ is the initial velocity (here, $u = 0$), $a$ is the acceleration, and $t$ is the time.

Once the body reaches a constant speed, it covers distance at that constant speed. The distance in this phase is simply: $s = v \cdot t,$ where $v$ is the constant speed and $t$ is the time spent at that speed.

In this problem, the train undergoes two phases:

1. Uniform acceleration from rest to $80 \text{ km/h}$ over time $t$.
2. Constant speed of $80 \text{ km/h}$ for time $3t$.

We are to find the average speed over the entire journey.

Step-by-Step Derivation:

Step 1: Convert units and define variables
The speed is given in $\text{km/h}$, and time is in hours (implied by the units). Let:

• $v_{\text{max}} = 80 \text{ km/h}$ (final speed after acceleration)
• $t$: time spent accelerating
• $3t$: time spent at constant speed

Step 2: Find acceleration during the first phase
The train starts from rest ($u = 0$) and reaches $v = 80 \text{ km/h}$ in time $t$. Using the equation: $v = u + a t \implies 80 = 0 + a t \implies a = \frac{80}{t} \text{ km/h}^2.$

Step 3: Calculate distance covered during acceleration ($s_1$)
Using $s = ut + \frac{1}{2} a t^2$, with $u = 0$: $s_1 = 0 \cdot t + \frac{1}{2} \cdot \frac{80}{t} \cdot t^2 = \frac{1}{2} \cdot 80 \cdot t = 40 t \text{ km}.$

Step 4: Calculate distance covered at constant speed ($s_2$)
The train moves at $80 \text{ km/h}$ for $3t$: $s_2 = 80 \cdot 3t = 240 t \text{ km}.$

Step 5: Compute total distance and total time
Total distance: $s_{\text{total}} = s_1 + s_2 = 40 t + 240 t = 280 t \text{ km}.$ Total time: $t_{\text{total}} = t + 3t = 4t.$

Step 6: Calculate average speed
$\text{Average Speed} = \frac{s_{\text{total}}}{t_{\text{total}}} = \frac{280 t}{4 t} = 70 \text{ km/h}.$ The $t$ cancels out, confirming the result is independent of $t$.

Common Traps & Exam Tip:

Trap 1: Incorrectly assuming average speed is the arithmetic mean of speeds.
Many students mistakenly compute $\frac{0 + 80}{2} = 40$ for the first phase and then average it with 80, leading to incorrect results. Average speed depends on distance and time, not just speeds.

Trap 2: Forgetting to convert units or misapplying equations of motion.
Ensure all units are consistent (here, km and hours are consistent). Also, remember that during acceleration, distance is not simply $v \cdot t$, but involves $\frac{1}{2} a t^2$.

Exam Tip:
Always break the motion into distinct phases (accelerated and uniform). Calculate distance and time for each phase separately, then sum them to find average speed. This structured approach minimizes errors.

Thus, the correct answer is A: 70.